The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.

[asy]
unitsize(0.8cm);
for (int i=0; i<7; ++i) {
for (int j=0; j<4; ++j) {
dot((i,j));
};}
label("$A$",(0,3),W);
label("$B$",(6,0),E);
label("$D$",(2,0),S);
label("$E$",(3.4,1.3),S);
dot((3.4,1.3));
label("$C$",(4,2),N);
draw((0,3)--(6,0),linewidth(0.7));
draw((2,0)--(4,2),linewidth(0.7));
[/asy]
Answer: Extend $\overline{DC}$ to $F$. Triangle $FAE$ and $DBE$ are similar with ratio $5:4$. Thus $AE=\frac{5AB}{9}$, $AB=\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5}$, and $AE=\frac{5(3\sqrt{5})}{9}=\boxed{\frac{5\sqrt{5}}{3}}$. [asy]
unitsize(0.8cm);
for (int i=0; i<7; ++i) {
for (int j=0; j<4; ++j) {
dot((i,j));
};}
label("$F$",(5,3),N);
label("$C$",(4,2),N);
draw((2,0)--(5,3)--(0,3)--(6,0)--cycle,linewidth(0.7));
label("$A$",(0,3),W);
label("$B$",(6,0),E);
label("$D$",(2,0),S);
label("$E$",(3.4,1.3),N);
dot((3.4,1.3));
label("$C$",(4,2),N);

[/asy]